slyte33 Posted November 30, 2009 Share Posted November 30, 2009 Basically I'd like to show 5 images within 2 lines; example: $images = 5; echo "<img src=>"; $images is set to 5, so echo the image 5 times I don't exactly need it for this approach, but the answer to it is all I need I thought foreach works, but I'm not sure how to do it. All help is appreciated, thanks Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/ Share on other sites More sharing options...
aeroswat Posted November 30, 2009 Share Posted November 30, 2009 i'm confused at what you are trying to do... will a simple for loop work for you? for($i=1;$i<=$images;i++) {echo "<img src=>";} Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/#findComment-968261 Share on other sites More sharing options...
slyte33 Posted November 30, 2009 Author Share Posted November 30, 2009 i'm confused at what you are trying to do... will a simple for loop work for you? for($i=1;$i<=$images;i++) {echo "<img src=>";} I'm unsure what this does and tried using it and got: Parse error: syntax error, unexpected T_INC, expecting ')' in /home/content/54/4743054/html/tbattle_findings.php on line 186 But from looking at it.. I see this is probably exactly what i'm looking for, just theres errors. Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/#findComment-968266 Share on other sites More sharing options...
aeroswat Posted November 30, 2009 Share Posted November 30, 2009 Is $images set like an integer or is it set with quotes like a string? also i left out a $ in the i++ should be $i++ Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/#findComment-968269 Share on other sites More sharing options...
premiso Posted November 30, 2009 Share Posted November 30, 2009 for is a loop that will basically loop and increment a variable for you: $images = 5; for ($i=1;$i<=$images;$i++) {echo "<img src=>";} There is a corrected version of the code. Basically it sets $i equal to 1 at the beginning and while $i <= $images (5) execute the code in the brackets, after executing increment $i by one and go back through the check....Hopefully I explained that process right lol. The correction to the original code as putting $ before the i++ Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/#findComment-968270 Share on other sites More sharing options...
slyte33 Posted November 30, 2009 Author Share Posted November 30, 2009 Ahh I see, it works fine, thank you Also, this would mean I could do this: echo "Image #$i<br>"; or if ($i == 5) { echo "This is image number 5"; } Thank you very much Link to comment https://forums.phpfreaks.com/topic/183437-display-x-images-based-on-a-number/#findComment-968274 Share on other sites More sharing options...
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